Let $R$ be the region enclosed by the polar curve $r(\theta)=3\sin(4\theta)$ where $\pi\leq \theta\leq \dfrac{5\pi}{4}$. $y$ $x$ $R$ $ 1$ $ 1$ Which integral represents the area of $R$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{4}\cdot\sin^2(4\theta)d\theta$ (Choice B) B $ \int_{\scriptsize\dfrac{\pi}{2}}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{2}\cdot\sin^2(4\theta)d\theta$ (Choice C) C $ \int_{\pi}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{2}\cdot\sin^2(4\theta)d\theta$ (Choice D) D $ \int_{\pi}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{4}\cdot\sin^2(4\theta)d\theta$
This is the formula for the area enclosed by a polar curve $r(\theta)$ between $\theta=\alpha$ and $\theta=\beta$ : $ \int_{\alpha}^{\beta}\dfrac{1}{2}\left(r(\theta)\right)^{2}d\theta$ Let's plug ${r(\theta)=3\sin(4\theta)}$, ${\alpha=\pi}$, and ${\beta=\dfrac{5\pi}{4}}$ into the formula and expand the parentheses: $\begin{aligned} &\phantom{=} \int_{\alpha}^{\beta}\dfrac{1}{2}\left({r(\theta)}\right)^{2}d\theta \\\\ &= \int_{{\pi}}^{{\scriptsize\dfrac{5\pi}{4}}}\dfrac{1}{2}\left({3\sin(4\theta)}\right)^{2}d\theta \\\\ &= \int_{\pi}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{1}{2}\cdot9\sin^2(4\theta)d\theta \\\\ &= \int_{\pi}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{2}\cdot\sin^2(4\theta)d\theta \end{aligned}$ In conclusion, this integral represents the area of region $R$ : $ \int_{\pi}^{\scriptsize\dfrac{5\pi}{4}}\dfrac{9}{2}\cdot\sin^2(4\theta)d\theta$